Hydraulic Engineering: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

Continuity & Bernoulli

Continuity: A₁v₁ = A₂v₂
Bernoulli: p₁ + ½ρv₁² + ρgz₁ = p₂ + ½ρv₂² + ρgz₂
Flow rate: Q = A·v
Circle area: A = πd²/4

Pipe friction & head loss

Reynolds number: Re = V·D/ν = ρ·V·D/μ
Flow regime: laminar below Re ≈ 2000, turbulent above Re ≈ 4000
Mean velocity: V = Q/A, A = π·D²/4
Darcy–Weisbach: h_f = f·(L/D)·V²/(2g)
Minor losses: h_m = ΣK·V²/(2g)
Typical K values: entrance 0.5, exit 1.0, 90° bend ≈ 0.3
Pressure drop: Δp = ρ·g·h_f
Two open reservoirs: Δz = (f·L/D + ΣK)·V²/(2g)
Free outlet: H = (1 + f·L/D + ΣK)·V²/(2g)

Pumps & pump power

Hydraulic power: P_h = ρ·g·Q·H = Q·Δp
Shaft power: P_shaft = P_h/η_pump
Electrical power: P_el = P_shaft/η_motor
Pump head: H = Δz + Δp/(ρ·g) + h_L
System losses: h_L = (f·L/D + ΣK)·V²/(2g)

Open-channel flow (Manning)

Manning equation: V = (1/n)·R^2/3·S^1/2
Flow rate: Q = A·V
Hydraulic radius: R = A/P (P = wetted perimeter, free surface excluded)
Rectangle: A = b·y, P = b + 2y
Trapezoid (side slope z:1): A = (b + z·y)·y, P = b + 2y·√(1 + z²)
Circular pipe, full or half full: R = D/4
Required slope: S = (n·Q/(A·R^2/3))²

2Practice problems

1.Hydrostatic forceMedium

A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 4 mH = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 1 mH = 2.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water drains from a large open tank through a sharp-edged orifice with a diameter of 25 mm. The water level is kept 2.5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².

d = 25 mmh = 2.5 mC_d = 0.62g = 9.81 m/s²

Q = L/s

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 40 kPa. Water leaves through an opening with a diameter of 40 mm, 1.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².

p_air = 40 kPa (gauge)h = 1.5 md = 40 mmρ = 1000 kg/m³g = 9.81 m/s²

v = m/s

5.Pipe friction & head lossMedium

Water flows at 80 L/s through a pipe 500 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 80 L/sL = 500 mD = 250 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 200 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.015) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each) and a valve (K = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 40 L/s?

Q = 40 L/sL = 200 mD = 150 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 56 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 50 L/sH = 56 mη_pump = 60%η_motor = 94%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 2.5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power P_shaft in kW.

Q = 75 L/sΔz = 5 mL = 100 mD = 200 mmf = 0.025ΣK = 2.5η = 70%

P_shaft = kW

9.Open-channel flow (Manning)Medium

A concrete storm sewer with an inside diameter of 900 mm flows half full. Manning n = 0.013 and the slope is S = 0.005. Find the flow rate Q in m³/s.

D = 900 mmn = 0.013S = 0.005

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal gravel channel with a bottom width of 2 m and side slopes of 1:1 (horizontal:vertical) must carry 17 m³/s in uniform flow at a depth of 1.7 m. Manning n = 0.025. Find the required bed slope S (in m/m).

Q = 17 m³/sb = 2 my = 1.7 mz = 1n = 0.025

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 196.2 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 4 + 2/2 = 5 m

A = H·b = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5·4 = 196.2 kN

Answer

F = 196.2 kN

2.Hydrostatic force

Answer: F = 72.84 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 1 + 2.5/2 = 2.25 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·2.25·3.75 = 72.84 kN

Answer

F = 72.84 kN

3.Continuity & Bernoulli

Answer: Q = 2.131 L/s

Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.

Ideal jet velocity

√(2gh) = √(2·9.81·2.5) = 7.004 m/s

Actual flow rate

The discharge coefficient accounts for the contraction of the jet and friction at the edge.

Q = C_d·A·√(2gh) = 0.62·0.0004909·7.004 = 0.002131 m³/s = 2.131 L/s

Answer

Q = 2.131 L/s

4.Continuity & Bernoulli

Answer: v = 10.46 m/s

Bernoulli from the surface to the jet

v = √(2·(40000/1000 + 9.81·1.5)) = 10.46 m/s

Answer

v = 10.46 m/s

5.Pipe friction & head loss

Answer: h_f = 5.415 m

Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss h_f = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses h_m = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = h_f + h_m, with the exit loss K = 1 included.

Mean velocity

A = π·0.25²/4 = 0.04909 m²

V = Q/A = 0.08/0.04909 = 1.63 m/s

Velocity head

V²/(2g) = 1.63²/(2·9.81) = 0.1354 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(500/0.25)·1.63²/(2·9.81) = 5.415 m

Answer

h_f = 5.415 m

6.Pipe friction & head loss

Answer: H = 6.294 m

Velocity

V = Q/A = 0.04/0.01767 = 2.264 m/s

V²/(2g) = 0.2611 m

Loss coefficients

f·L/D = 0.015·200/0.15 = 20

ΣK = 0.5 + 2·0.3 + 2 = 3.1

Energy equation

The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.

H = (1 + 20 + 3.1)·0.2611 = 6.294 m

Answer

H = 6.294 m

7.Pumps & pump power

Answer: P_el = 48.7 kW

Pump power: the power given to the water is the hydraulic power P_h = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: P_shaft = P_h/η_pump, and the motor draws P_el = P_shaft/η_motor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

P_h = ρ·g·Q·H = 1000·9.81·0.05·56 = 27.47 kW

Shaft power

P_shaft = 27.47/0.6 = 45.78 kW

Electrical power

P_el = 45.78/0.94 = 48.7 kW

Answer

P_el = 48.7 kW

8.Pumps & pump power

Answer: P_shaft = 9.835 kW

Velocity

V = Q/A = 0.075/0.03142 = 2.387 m/s

V²/(2g) = 0.2905 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (12.5 + 2.5)·0.2905 = 4.357 m

Pump head

H = Δz + h_L = 5 + 4.357 = 9.357 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.075·9.357/0.7 = 9.835 kW

Answer

P_shaft = 9.835 kW

9.Open-channel flow (Manning)

Answer: Q = 0.64 m³/s

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R^2/3·S^1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Geometry

A = 0.3181 m², R = D/4 = 0.225 m

Velocity

V = (1/0.013)·0.225^2/3·0.005^1/2 = 2.012 m/s

Flow rate

Q = A·V = 0.3181·2.012 = 0.64 m³/s

Answer

Q = 0.64 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.005074 m/m

Geometry

A = (b + z·y)·y = (2 + 1·1.7)·1.7 = 6.29 m²

P = b + 2y·√(1 + z²) = 2 + 2·1.7·√(1 + 1²) = 6.808 m

R = A/P = 0.9239 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.025·17/(6.29·0.9486) = 0.07123

S = 0.07123² = 0.005074

Answer

S = 0.005074

That is a fall of 5.074 m per kilometre.

ExamProblems · examproblems.com/courses/hydraulic-engineering